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Suppose we have a quantum state, well described by its time-independent wave function Psi. And we have a well-defined Hermitian (self-adjoint) operator $A$. We successfully evaluate the expectation value of the operator $A$. Next we derive the general formula for the higher moments of $A$ (i.e. the expectation value of $A^n$ for $n=2,3,4\ldots $).

In this situation, is it permitted to regard each of the $A^n$ for $n=1,2,3,\ldots$ as a proper operator by itself?

For example, should every $A^n$ have a positive variance and other statistical properties (as long as we restrict ourselves to the state $\Psi$)?

Can one make linear combinations of different powers to construct a new operator, e.g. $B = A + A^2$?

Is it allowed to construct new operators acting on $\Psi$, that are defined in terms of their series expansion in $A^n$? For example, $D = \exp(CA)$ where $C$ is a constant?

Qmechanic
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M. Wind
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    $|\psi(t)\rangle = e^{-\frac{i\hat H}{\hbar}t}|\psi(0)\rangle$: http://en.wikipedia.org/wiki/Hamiltonian_%28quantum_mechanics%29#Schr.C3.B6dinger_equation – Alfred Centauri Jun 12 '14 at 23:46

3 Answers3

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Most of your question is unclear to me, but I the answer to what I think is the core of your question is yes:

  • For any hermitian operator $A$ and any well-behaved-enough function $f:\mathbb R\to\mathbb R$, it is possible to construct a new operator $f(A)$ which acts in essentially all important respects as the action of $f$ on $A$.

There are different ways to construct $f(A)$, and they depend on exactly what $f$ is, what "well-behaved-enough" means, and how nice $A$ is. In general, the keyword to search for is function of an operator.

For example, if $f$ is analytic in a large enough region - one that includes all the spectrum of $A$, for example, then you can define it as $$ f(A)=\sum_{n=0}^\infty \frac{f^{(n)}(a_0)}{n!}(A-a_0)^n. $$

Note, in particular, that this includes functions of the form $A+A^2$, which are perfectly allowed. If $A$ is a linear operator then so is $A^2$, and adding two linear operators is bread and butter in linear algebra. There are a few caveats - for example, if the domain of $A$ is smaller than the Hilbert space then the domain of $A^2$ will typically be smaller, so the sum only makes sense in that restricted domain - but this is only the standard measure of care one needs to take when infinite-dimensional spaces are involved.

Alternatively, if $A$ has an eigenvector expansion as $A=\sum_k a_k|a_k\rangle\langle a_k|$, then you can define $$f(A)=\sum_k f(a_k)|a_k\rangle\langle a_k|.$$ If everything behaves well, then both definitions will match.

Finally, note that one should keep an eye on the dimensional analysis of the whole thing. If $A=x$ has dimensions of position, then it does not make sense to add $x+x^2$, any more than it does to do this in classical mechanics. This is particularly the case with, say, exponentials of operators, like the displacement operator $$ e^{ix_0\hat p}=\sum_{n=0}^\infty i^n\frac{x_0^n}{n!}\hat p^n $$ which only makes sense in units where $\hbar=1$. (Otherwise, you need to add in the $\hbar$ explicitly, as $\exp(ix_0\hat p/\hbar$.)

Emilio Pisanty
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  • There is another possibility to define $f(A)$, which afaik needs less requirements on $f$ and $A$. Specifically, it works for Borel-measurable functions $f: \mathbb R \rightarrow \mathbb C$ and self-adjoint operators $A$. $f(A)$ is then given by the Helffer-Sjöstrand formula: $f(A)=\frac{1}{\pi}\int_{\mathbb C} \frac{\partial \tilde f}{\partial \bar z}(z)(A-z)^{-1}\mathrm{d}x \mathrm{d}y$. Here, $\tilde f$ is the almost analytic continuation of $f$, $z=x + \mathrm{i}y$ and $(A-z)^{-1}$ the resolvent of $A$. – Simeon Carstens Jun 13 '14 at 08:52
  • It's been years that I learned about this and I have to admit I never really understood completely. So to the more mathematically inclined people here: please correct me if I'm mistaken in the above comment. – Simeon Carstens Jun 13 '14 at 08:56
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    @Emilio Pisanty Hi Emilio. Unfortunately your recipe based on Taylor series generally fails unless $A$ is bounded (which is quite uncommon in general QM). If restricting to very particular cases of $f$ like the exponential, it works applying bot sides to very particular vectors called "analytic vectors". The only always working procedure is the spectral one: $f(A) := \int_{\sigma(A)} f(\lambda) dP^{(A)}(\lambda)$ for every Borel measurable function $f: \sigma(A) \to \mathbb C$, where $P^{(A)}$ is the spectral measure of $A$ (supposed to be self-adjoint or at least normal). – Valter Moretti Jun 13 '14 at 10:00
  • @V.Moretti thanks for the precision. I was deliberately hazy (hence "well-behaved-enough") partly because I don't know the precise results off the top of my head but also to avoid dizzying the OP with spectral measures. But point taken. – Emilio Pisanty Jun 14 '14 at 00:49
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Sure. Anything that maps one state to another is an operator. If $A$ satisfies this definition, namely that when applied to a state it gives you a state, then so does repeated application of $A$.

For example, suppose you have a set of quantum states $\lvert i\rangle$ for various values of $i$, parametrized so that $A\lvert i\rangle = \lvert i+1\rangle$. Then

$$A^2\lvert i\rangle = A(A\lvert i\rangle) = A\lvert i+1\rangle = \lvert i+2\rangle$$

Hopefully you can see how this generalizes, so that $A^2$ is the operator that takes $\lvert i\rangle\to\lvert i+2\rangle$.

And yes, you can generalize this to construct an operator as a power series of other operators. This is how the exponential of an operator is defined, for example.

David Z
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  • Thank you very much for your clear reply! I am worried that this great freedom to create new operators in terms of linear combinations of other operators can lead to odd results, that are in conflict with statistical rules. – M. Wind Jun 13 '14 at 13:07
  • In fact I have just posted a new thread "Significance of an operator with a negative variance". It is an example of an operator A with fairly normal moments evaluated in some state. From this a new operator B is constructed which has very odd properties. – M. Wind Jun 13 '14 at 14:28
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For a continuous (linear) operator $A:H\to H$, aka. bounded operator, it is always possible to construct well-defined powers $A^2$, $A^3$, $\ldots.$ Here $H$ denotes a (complex) Hilbert space.

However, the situation changes drastically for a general unbounded operator $A$. Unbounded operators often appear in quantum mechanics, see e.g. this, this and this Phys.SE posts.

The domain $D(A)\subsetneq H$ of an unbounded operator is never the full Hilbert space! Therefore if the image ${\rm Im}(A)\subseteq H$ is not a subset of the domain $D(A)$, then the square operator $A^2$ does not necessarily make sense on the full domain $D(A)$ of $A$. In other words, the domain $D(A^2)$ of the square operator $A^2$ is in general different from the domain of $A$! Similar for higher powers of $A$.

The topic of unbounded operators is a huge subject in functional analysis, which is impossible to cover in a single Phys.SE post. For a student of operator theory, the natural next couple of questions to ponder is:

  1. Is it possible to extend a domain $D$ of an unbounded operator $A$?

  2. If yes, is there a natural way to partially order the set of all possible domains of an unbounded operator $A$?

  3. Is there a canonical choice of a domain for an unbounded operator $A$?

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